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Java 8 stream max() function argument type Comparator vs Comparable
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Data science time! April 2019 and salary with experience
The Ask Question Wizard is Live!Java 8 stream's .min() and .max(): why does this compile?How do I compare strings in Java?Comparing Java enum members: == or equals()?Best practice for compareTo() when argument must be typed of super classIs there a concise way to iterate over a stream with indices in Java 8?Custom thread pool in Java 8 parallel streamHow to Convert a Java 8 Stream to an Array?Convert Iterable to Stream using Java 8 JDKWhy are Java Streams once-off?How to sum a list of integers with java streams?Java Generics Clarification( Constraining T to a type, while using Comparable)
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I wrote some simple code like below. This class works fine without any errors.
public class Test
public static void main(String[] args)
List<Integer> intList = IntStream.of(1,2,3,4,5,6,7,8,9,10).boxed().collect(Collectors.toList());
int value = intList.stream().max(Integer::compareTo).get();
//int value = intList.stream().max(<Comparator<? super T> comparator type should pass here>).get();
System.out.println("value :"+value);
As the code comment shows the max()
method should pass an argument of type Comparator<? super Integer>
.
But Integer::compareTo
implements Comparable
interface - not Comparator
.
public final class Integer extends Number implements Comparable<Integer>
public int compareTo(Integer anotherInteger)
return compare(this.value, anotherInteger.value);
How can this work? The max()
method says it needs a Comparator
argument, but it works with Comparable
argument.
I know I have misunderstood something, but I do now know what. Can someone please explain?
java java-8 java-stream
add a comment |
I wrote some simple code like below. This class works fine without any errors.
public class Test
public static void main(String[] args)
List<Integer> intList = IntStream.of(1,2,3,4,5,6,7,8,9,10).boxed().collect(Collectors.toList());
int value = intList.stream().max(Integer::compareTo).get();
//int value = intList.stream().max(<Comparator<? super T> comparator type should pass here>).get();
System.out.println("value :"+value);
As the code comment shows the max()
method should pass an argument of type Comparator<? super Integer>
.
But Integer::compareTo
implements Comparable
interface - not Comparator
.
public final class Integer extends Number implements Comparable<Integer>
public int compareTo(Integer anotherInteger)
return compare(this.value, anotherInteger.value);
How can this work? The max()
method says it needs a Comparator
argument, but it works with Comparable
argument.
I know I have misunderstood something, but I do now know what. Can someone please explain?
java java-8 java-stream
4
Integer::compareTo
does not return a Comparable - it is the short definition for: "Please compiler generate a matching implementation for the type that is needed (in this caseComparator
) and map the arguments to the specified function." In this case the function requires two "arguments" (this
and the one parameter ofcompareTo
) and the Comparator provides two arguments -> works.
– Robert
Apr 15 at 18:37
6
""Please compiler, generate..." ... compilers always respond best to politeness and courtesy :-)
– scottb
Apr 15 at 18:45
1
Related: stackoverflow.com/questions/22561614/…
– Oleksandr
Apr 15 at 22:27
add a comment |
I wrote some simple code like below. This class works fine without any errors.
public class Test
public static void main(String[] args)
List<Integer> intList = IntStream.of(1,2,3,4,5,6,7,8,9,10).boxed().collect(Collectors.toList());
int value = intList.stream().max(Integer::compareTo).get();
//int value = intList.stream().max(<Comparator<? super T> comparator type should pass here>).get();
System.out.println("value :"+value);
As the code comment shows the max()
method should pass an argument of type Comparator<? super Integer>
.
But Integer::compareTo
implements Comparable
interface - not Comparator
.
public final class Integer extends Number implements Comparable<Integer>
public int compareTo(Integer anotherInteger)
return compare(this.value, anotherInteger.value);
How can this work? The max()
method says it needs a Comparator
argument, but it works with Comparable
argument.
I know I have misunderstood something, but I do now know what. Can someone please explain?
java java-8 java-stream
I wrote some simple code like below. This class works fine without any errors.
public class Test
public static void main(String[] args)
List<Integer> intList = IntStream.of(1,2,3,4,5,6,7,8,9,10).boxed().collect(Collectors.toList());
int value = intList.stream().max(Integer::compareTo).get();
//int value = intList.stream().max(<Comparator<? super T> comparator type should pass here>).get();
System.out.println("value :"+value);
As the code comment shows the max()
method should pass an argument of type Comparator<? super Integer>
.
But Integer::compareTo
implements Comparable
interface - not Comparator
.
public final class Integer extends Number implements Comparable<Integer>
public int compareTo(Integer anotherInteger)
return compare(this.value, anotherInteger.value);
How can this work? The max()
method says it needs a Comparator
argument, but it works with Comparable
argument.
I know I have misunderstood something, but I do now know what. Can someone please explain?
java java-8 java-stream
java java-8 java-stream
edited Apr 15 at 19:37
MyStackRunnethOver
1,009919
1,009919
asked Apr 15 at 16:57
ShalikaShalika
4181618
4181618
4
Integer::compareTo
does not return a Comparable - it is the short definition for: "Please compiler generate a matching implementation for the type that is needed (in this caseComparator
) and map the arguments to the specified function." In this case the function requires two "arguments" (this
and the one parameter ofcompareTo
) and the Comparator provides two arguments -> works.
– Robert
Apr 15 at 18:37
6
""Please compiler, generate..." ... compilers always respond best to politeness and courtesy :-)
– scottb
Apr 15 at 18:45
1
Related: stackoverflow.com/questions/22561614/…
– Oleksandr
Apr 15 at 22:27
add a comment |
4
Integer::compareTo
does not return a Comparable - it is the short definition for: "Please compiler generate a matching implementation for the type that is needed (in this caseComparator
) and map the arguments to the specified function." In this case the function requires two "arguments" (this
and the one parameter ofcompareTo
) and the Comparator provides two arguments -> works.
– Robert
Apr 15 at 18:37
6
""Please compiler, generate..." ... compilers always respond best to politeness and courtesy :-)
– scottb
Apr 15 at 18:45
1
Related: stackoverflow.com/questions/22561614/…
– Oleksandr
Apr 15 at 22:27
4
4
Integer::compareTo
does not return a Comparable - it is the short definition for: "Please compiler generate a matching implementation for the type that is needed (in this case Comparator
) and map the arguments to the specified function." In this case the function requires two "arguments" (this
and the one parameter of compareTo
) and the Comparator provides two arguments -> works.– Robert
Apr 15 at 18:37
Integer::compareTo
does not return a Comparable - it is the short definition for: "Please compiler generate a matching implementation for the type that is needed (in this case Comparator
) and map the arguments to the specified function." In this case the function requires two "arguments" (this
and the one parameter of compareTo
) and the Comparator provides two arguments -> works.– Robert
Apr 15 at 18:37
6
6
""Please compiler, generate..." ... compilers always respond best to politeness and courtesy :-)
– scottb
Apr 15 at 18:45
""Please compiler, generate..." ... compilers always respond best to politeness and courtesy :-)
– scottb
Apr 15 at 18:45
1
1
Related: stackoverflow.com/questions/22561614/…
– Oleksandr
Apr 15 at 22:27
Related: stackoverflow.com/questions/22561614/…
– Oleksandr
Apr 15 at 22:27
add a comment |
4 Answers
4
active
oldest
votes
int value = intList.stream().max(Integer::compareTo).get();
The above snippet of code is logically equivalent to the following:
int value = intList.stream().max((a, b) -> a.compareTo(b)).get();
Which is also logically equivalent to the following:
int value = intList.stream().max(new Comparator<Integer>()
@Override
public int compare(Integer a, Integer b)
return a.compareTo(b);
).get();
Comparator
is a functional interface and can be used as a lambda or method reference, which is why your code compiles and executes successfully.
I recommend reading Oracle's tutorial on Method References (they use an example where two objects are compared) as well as the Java Language Specification on §15.13. Method Reference Expressions to understand why this works.
9
Although it's absolutely correct, it doesn't answer "How can this work?"
– Andrew Tobilko
Apr 15 at 17:18
add a comment |
I can relate to your confusion.
We've got a Comparator
's method which declares two parameters
int compare(T o1, T o2);
and we've got an Integer
's method which takes one parameter
int compareTo(Integer anotherInteger)
How on earth does Integer::compareTo
get resolved to a Comparator
instance?
When a method reference points to an instance method, the parser can look for methods with arity n-1
(n
is the expected number of parameters).
Here's an excerpt from the JLS on how applicable methods are identified. I will drop the first part about parsing the expression preceding the ::
token.
Second, given a targeted function type with
n
parameters, a set of potentially applicable methods is identified:
If the method reference expression has the form
ReferenceType :: [TypeArguments] Identifier
, then the potentially applicable methods are:
the member methods of the type to search that would be potentially applicable (§15.12.2.1) for a method invocation which names Identifier, has arity n, has type arguments TypeArguments, and appears in the same class as the method reference expression; plus
the member methods of the type to search that would be potentially applicable for a method invocation which names
Identifier
, has arity n-1, has type arguments TypeArguments, and appears in the same class as the method reference expression.
Two different arities,
n
andn-1
, are considered, to account for the possibility that this form refers to either a static method or an instance method.
...
A method reference expression of the form
ReferenceType :: [TypeArguments] Identifier
can be interpreted in different ways. IfIdentifier
refers to an instance method, then the implicit lambda expression has an extra parameter compared to ifIdentifier
refers to a static method.
https://docs.oracle.com/javase/specs/jls/se12/html/jls-15.html#jls-15.13.1
If we were to write an implicit lambda expression from that method reference, the first (implicit) parameter would be an instance to call the method on, the second (explicit) parameter would be an argument to pass in the method.
(implicitParam, anotherInteger) -> implicitParam.compareTo(anotherInteger)
Note that a method reference differs from a lambda expression, even though the former can be easily transformed into the latter. A lambda expression needs to be desugared into a new method, while a method reference usually requires only loading a corresponding constant method handle.
Integer::compareTo
implementsComparable
interface - notComparator
.
Integer::compareTo
as an expression doesn't implement any interface. However, it can refer to/represent different functional types, one of which is Comparator<Integer>
.
Comparator<Integer> a = Integer::compareTo;
BiFunction<Integer, Integer, Integer> b = Integer::compareTo;
ToIntBiFunction<Integer, Integer> c = Integer::compareTo;
add a comment |
Integer
implements Comparable
by overriding compareTo
.
That overriden compareTo
, however, can be used in a way that satisfies and implements the Comparator
interface.
In its usage here
int value = intList.stream().max(Integer::compareTo).get();
it's translated to something like
int value = intList.stream().max(new Comparator<Integer>()
@Override
public int compare(Integer o1, Integer o2)
return o1.compareTo(o2);
).get();
A method reference (or lambda expression) must satisfy the signature of the corresponding functional interface's single abstract method and, in this case (Comparator
), compareTo
does.
The idea is that max
expects a Comparator
and its compare
method expects two Integer
objects. Integer::compareTo
can satisfy those expectations because it also expects two Integer
objects. The first is its receiver (the instance on which the method is to be called) and the second is the argument. With the new Java 8 syntax, the compiler translates one style to the other.
(compareTo
also returns an int
as required by Comparator#compare
.)
1
how doesInteger::compareTo
satisfy the signature ofComparator#compare
?compareTo(Integer)
vscompare(Integer, Integer)
?
– Andrew Tobilko
Apr 15 at 21:30
1
I disagree with "the compiler translates one style to the other". I think the compiler may validate a syntactic construct, but not translate it into another one. I am sure the compiler doesn't turn a method reference in a lambda. I didn't downvote your answer, btw.
– Andrew Tobilko
Apr 16 at 7:02
add a comment |
First trick: all instance methods actually take 1 additional implicit argument, the one you refer to as this
in method body. E.g.:
public final class Integer extends Number implements Comparable<Integer>
public int compareTo(/* Integer this, */ Integer anotherInteger)
return compare(this.value, anotherInteger.value);
Integer a = 10, b = 100;
int compareResult = a.compareTo(b);
// this actually 'compiles' to Integer#compareTo(this = a, anotherInteger = b)
Second trick: Java compiler can "transform" the signature of a method reference to some functional interface, if the number and types of arguments (including this
) satisfy:
interface MyInterface
int foo(Integer bar, Integer baz);
Integer a = 100, b = 1000;
int result1 = ((Comparator<Integer>) Integer::compareTo).compare(a, b);
int result2 = ((BiFunction<Integer, Integer, Integer>) Integer::compareTo).apply(a, b);
int result3 = ((MyInterface) Integer::compareTo).foo(a, b);
// result1 == result2 == result3
As you can see class Integer
implements none of Comparator
, BiFunction
or a random MyInterface
, but that doesn't stop you from casting the Integer::compareTo
method reference as those interfaces.
add a comment |
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4 Answers
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active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
int value = intList.stream().max(Integer::compareTo).get();
The above snippet of code is logically equivalent to the following:
int value = intList.stream().max((a, b) -> a.compareTo(b)).get();
Which is also logically equivalent to the following:
int value = intList.stream().max(new Comparator<Integer>()
@Override
public int compare(Integer a, Integer b)
return a.compareTo(b);
).get();
Comparator
is a functional interface and can be used as a lambda or method reference, which is why your code compiles and executes successfully.
I recommend reading Oracle's tutorial on Method References (they use an example where two objects are compared) as well as the Java Language Specification on §15.13. Method Reference Expressions to understand why this works.
9
Although it's absolutely correct, it doesn't answer "How can this work?"
– Andrew Tobilko
Apr 15 at 17:18
add a comment |
int value = intList.stream().max(Integer::compareTo).get();
The above snippet of code is logically equivalent to the following:
int value = intList.stream().max((a, b) -> a.compareTo(b)).get();
Which is also logically equivalent to the following:
int value = intList.stream().max(new Comparator<Integer>()
@Override
public int compare(Integer a, Integer b)
return a.compareTo(b);
).get();
Comparator
is a functional interface and can be used as a lambda or method reference, which is why your code compiles and executes successfully.
I recommend reading Oracle's tutorial on Method References (they use an example where two objects are compared) as well as the Java Language Specification on §15.13. Method Reference Expressions to understand why this works.
9
Although it's absolutely correct, it doesn't answer "How can this work?"
– Andrew Tobilko
Apr 15 at 17:18
add a comment |
int value = intList.stream().max(Integer::compareTo).get();
The above snippet of code is logically equivalent to the following:
int value = intList.stream().max((a, b) -> a.compareTo(b)).get();
Which is also logically equivalent to the following:
int value = intList.stream().max(new Comparator<Integer>()
@Override
public int compare(Integer a, Integer b)
return a.compareTo(b);
).get();
Comparator
is a functional interface and can be used as a lambda or method reference, which is why your code compiles and executes successfully.
I recommend reading Oracle's tutorial on Method References (they use an example where two objects are compared) as well as the Java Language Specification on §15.13. Method Reference Expressions to understand why this works.
int value = intList.stream().max(Integer::compareTo).get();
The above snippet of code is logically equivalent to the following:
int value = intList.stream().max((a, b) -> a.compareTo(b)).get();
Which is also logically equivalent to the following:
int value = intList.stream().max(new Comparator<Integer>()
@Override
public int compare(Integer a, Integer b)
return a.compareTo(b);
).get();
Comparator
is a functional interface and can be used as a lambda or method reference, which is why your code compiles and executes successfully.
I recommend reading Oracle's tutorial on Method References (they use an example where two objects are compared) as well as the Java Language Specification on §15.13. Method Reference Expressions to understand why this works.
edited Apr 15 at 17:33
answered Apr 15 at 17:02
Jacob G.Jacob G.
16.9k52466
16.9k52466
9
Although it's absolutely correct, it doesn't answer "How can this work?"
– Andrew Tobilko
Apr 15 at 17:18
add a comment |
9
Although it's absolutely correct, it doesn't answer "How can this work?"
– Andrew Tobilko
Apr 15 at 17:18
9
9
Although it's absolutely correct, it doesn't answer "How can this work?"
– Andrew Tobilko
Apr 15 at 17:18
Although it's absolutely correct, it doesn't answer "How can this work?"
– Andrew Tobilko
Apr 15 at 17:18
add a comment |
I can relate to your confusion.
We've got a Comparator
's method which declares two parameters
int compare(T o1, T o2);
and we've got an Integer
's method which takes one parameter
int compareTo(Integer anotherInteger)
How on earth does Integer::compareTo
get resolved to a Comparator
instance?
When a method reference points to an instance method, the parser can look for methods with arity n-1
(n
is the expected number of parameters).
Here's an excerpt from the JLS on how applicable methods are identified. I will drop the first part about parsing the expression preceding the ::
token.
Second, given a targeted function type with
n
parameters, a set of potentially applicable methods is identified:
If the method reference expression has the form
ReferenceType :: [TypeArguments] Identifier
, then the potentially applicable methods are:
the member methods of the type to search that would be potentially applicable (§15.12.2.1) for a method invocation which names Identifier, has arity n, has type arguments TypeArguments, and appears in the same class as the method reference expression; plus
the member methods of the type to search that would be potentially applicable for a method invocation which names
Identifier
, has arity n-1, has type arguments TypeArguments, and appears in the same class as the method reference expression.
Two different arities,
n
andn-1
, are considered, to account for the possibility that this form refers to either a static method or an instance method.
...
A method reference expression of the form
ReferenceType :: [TypeArguments] Identifier
can be interpreted in different ways. IfIdentifier
refers to an instance method, then the implicit lambda expression has an extra parameter compared to ifIdentifier
refers to a static method.
https://docs.oracle.com/javase/specs/jls/se12/html/jls-15.html#jls-15.13.1
If we were to write an implicit lambda expression from that method reference, the first (implicit) parameter would be an instance to call the method on, the second (explicit) parameter would be an argument to pass in the method.
(implicitParam, anotherInteger) -> implicitParam.compareTo(anotherInteger)
Note that a method reference differs from a lambda expression, even though the former can be easily transformed into the latter. A lambda expression needs to be desugared into a new method, while a method reference usually requires only loading a corresponding constant method handle.
Integer::compareTo
implementsComparable
interface - notComparator
.
Integer::compareTo
as an expression doesn't implement any interface. However, it can refer to/represent different functional types, one of which is Comparator<Integer>
.
Comparator<Integer> a = Integer::compareTo;
BiFunction<Integer, Integer, Integer> b = Integer::compareTo;
ToIntBiFunction<Integer, Integer> c = Integer::compareTo;
add a comment |
I can relate to your confusion.
We've got a Comparator
's method which declares two parameters
int compare(T o1, T o2);
and we've got an Integer
's method which takes one parameter
int compareTo(Integer anotherInteger)
How on earth does Integer::compareTo
get resolved to a Comparator
instance?
When a method reference points to an instance method, the parser can look for methods with arity n-1
(n
is the expected number of parameters).
Here's an excerpt from the JLS on how applicable methods are identified. I will drop the first part about parsing the expression preceding the ::
token.
Second, given a targeted function type with
n
parameters, a set of potentially applicable methods is identified:
If the method reference expression has the form
ReferenceType :: [TypeArguments] Identifier
, then the potentially applicable methods are:
the member methods of the type to search that would be potentially applicable (§15.12.2.1) for a method invocation which names Identifier, has arity n, has type arguments TypeArguments, and appears in the same class as the method reference expression; plus
the member methods of the type to search that would be potentially applicable for a method invocation which names
Identifier
, has arity n-1, has type arguments TypeArguments, and appears in the same class as the method reference expression.
Two different arities,
n
andn-1
, are considered, to account for the possibility that this form refers to either a static method or an instance method.
...
A method reference expression of the form
ReferenceType :: [TypeArguments] Identifier
can be interpreted in different ways. IfIdentifier
refers to an instance method, then the implicit lambda expression has an extra parameter compared to ifIdentifier
refers to a static method.
https://docs.oracle.com/javase/specs/jls/se12/html/jls-15.html#jls-15.13.1
If we were to write an implicit lambda expression from that method reference, the first (implicit) parameter would be an instance to call the method on, the second (explicit) parameter would be an argument to pass in the method.
(implicitParam, anotherInteger) -> implicitParam.compareTo(anotherInteger)
Note that a method reference differs from a lambda expression, even though the former can be easily transformed into the latter. A lambda expression needs to be desugared into a new method, while a method reference usually requires only loading a corresponding constant method handle.
Integer::compareTo
implementsComparable
interface - notComparator
.
Integer::compareTo
as an expression doesn't implement any interface. However, it can refer to/represent different functional types, one of which is Comparator<Integer>
.
Comparator<Integer> a = Integer::compareTo;
BiFunction<Integer, Integer, Integer> b = Integer::compareTo;
ToIntBiFunction<Integer, Integer> c = Integer::compareTo;
add a comment |
I can relate to your confusion.
We've got a Comparator
's method which declares two parameters
int compare(T o1, T o2);
and we've got an Integer
's method which takes one parameter
int compareTo(Integer anotherInteger)
How on earth does Integer::compareTo
get resolved to a Comparator
instance?
When a method reference points to an instance method, the parser can look for methods with arity n-1
(n
is the expected number of parameters).
Here's an excerpt from the JLS on how applicable methods are identified. I will drop the first part about parsing the expression preceding the ::
token.
Second, given a targeted function type with
n
parameters, a set of potentially applicable methods is identified:
If the method reference expression has the form
ReferenceType :: [TypeArguments] Identifier
, then the potentially applicable methods are:
the member methods of the type to search that would be potentially applicable (§15.12.2.1) for a method invocation which names Identifier, has arity n, has type arguments TypeArguments, and appears in the same class as the method reference expression; plus
the member methods of the type to search that would be potentially applicable for a method invocation which names
Identifier
, has arity n-1, has type arguments TypeArguments, and appears in the same class as the method reference expression.
Two different arities,
n
andn-1
, are considered, to account for the possibility that this form refers to either a static method or an instance method.
...
A method reference expression of the form
ReferenceType :: [TypeArguments] Identifier
can be interpreted in different ways. IfIdentifier
refers to an instance method, then the implicit lambda expression has an extra parameter compared to ifIdentifier
refers to a static method.
https://docs.oracle.com/javase/specs/jls/se12/html/jls-15.html#jls-15.13.1
If we were to write an implicit lambda expression from that method reference, the first (implicit) parameter would be an instance to call the method on, the second (explicit) parameter would be an argument to pass in the method.
(implicitParam, anotherInteger) -> implicitParam.compareTo(anotherInteger)
Note that a method reference differs from a lambda expression, even though the former can be easily transformed into the latter. A lambda expression needs to be desugared into a new method, while a method reference usually requires only loading a corresponding constant method handle.
Integer::compareTo
implementsComparable
interface - notComparator
.
Integer::compareTo
as an expression doesn't implement any interface. However, it can refer to/represent different functional types, one of which is Comparator<Integer>
.
Comparator<Integer> a = Integer::compareTo;
BiFunction<Integer, Integer, Integer> b = Integer::compareTo;
ToIntBiFunction<Integer, Integer> c = Integer::compareTo;
I can relate to your confusion.
We've got a Comparator
's method which declares two parameters
int compare(T o1, T o2);
and we've got an Integer
's method which takes one parameter
int compareTo(Integer anotherInteger)
How on earth does Integer::compareTo
get resolved to a Comparator
instance?
When a method reference points to an instance method, the parser can look for methods with arity n-1
(n
is the expected number of parameters).
Here's an excerpt from the JLS on how applicable methods are identified. I will drop the first part about parsing the expression preceding the ::
token.
Second, given a targeted function type with
n
parameters, a set of potentially applicable methods is identified:
If the method reference expression has the form
ReferenceType :: [TypeArguments] Identifier
, then the potentially applicable methods are:
the member methods of the type to search that would be potentially applicable (§15.12.2.1) for a method invocation which names Identifier, has arity n, has type arguments TypeArguments, and appears in the same class as the method reference expression; plus
the member methods of the type to search that would be potentially applicable for a method invocation which names
Identifier
, has arity n-1, has type arguments TypeArguments, and appears in the same class as the method reference expression.
Two different arities,
n
andn-1
, are considered, to account for the possibility that this form refers to either a static method or an instance method.
...
A method reference expression of the form
ReferenceType :: [TypeArguments] Identifier
can be interpreted in different ways. IfIdentifier
refers to an instance method, then the implicit lambda expression has an extra parameter compared to ifIdentifier
refers to a static method.
https://docs.oracle.com/javase/specs/jls/se12/html/jls-15.html#jls-15.13.1
If we were to write an implicit lambda expression from that method reference, the first (implicit) parameter would be an instance to call the method on, the second (explicit) parameter would be an argument to pass in the method.
(implicitParam, anotherInteger) -> implicitParam.compareTo(anotherInteger)
Note that a method reference differs from a lambda expression, even though the former can be easily transformed into the latter. A lambda expression needs to be desugared into a new method, while a method reference usually requires only loading a corresponding constant method handle.
Integer::compareTo
implementsComparable
interface - notComparator
.
Integer::compareTo
as an expression doesn't implement any interface. However, it can refer to/represent different functional types, one of which is Comparator<Integer>
.
Comparator<Integer> a = Integer::compareTo;
BiFunction<Integer, Integer, Integer> b = Integer::compareTo;
ToIntBiFunction<Integer, Integer> c = Integer::compareTo;
edited Apr 16 at 7:10
answered Apr 15 at 17:51
Andrew TobilkoAndrew Tobilko
28.9k104592
28.9k104592
add a comment |
add a comment |
Integer
implements Comparable
by overriding compareTo
.
That overriden compareTo
, however, can be used in a way that satisfies and implements the Comparator
interface.
In its usage here
int value = intList.stream().max(Integer::compareTo).get();
it's translated to something like
int value = intList.stream().max(new Comparator<Integer>()
@Override
public int compare(Integer o1, Integer o2)
return o1.compareTo(o2);
).get();
A method reference (or lambda expression) must satisfy the signature of the corresponding functional interface's single abstract method and, in this case (Comparator
), compareTo
does.
The idea is that max
expects a Comparator
and its compare
method expects two Integer
objects. Integer::compareTo
can satisfy those expectations because it also expects two Integer
objects. The first is its receiver (the instance on which the method is to be called) and the second is the argument. With the new Java 8 syntax, the compiler translates one style to the other.
(compareTo
also returns an int
as required by Comparator#compare
.)
1
how doesInteger::compareTo
satisfy the signature ofComparator#compare
?compareTo(Integer)
vscompare(Integer, Integer)
?
– Andrew Tobilko
Apr 15 at 21:30
1
I disagree with "the compiler translates one style to the other". I think the compiler may validate a syntactic construct, but not translate it into another one. I am sure the compiler doesn't turn a method reference in a lambda. I didn't downvote your answer, btw.
– Andrew Tobilko
Apr 16 at 7:02
add a comment |
Integer
implements Comparable
by overriding compareTo
.
That overriden compareTo
, however, can be used in a way that satisfies and implements the Comparator
interface.
In its usage here
int value = intList.stream().max(Integer::compareTo).get();
it's translated to something like
int value = intList.stream().max(new Comparator<Integer>()
@Override
public int compare(Integer o1, Integer o2)
return o1.compareTo(o2);
).get();
A method reference (or lambda expression) must satisfy the signature of the corresponding functional interface's single abstract method and, in this case (Comparator
), compareTo
does.
The idea is that max
expects a Comparator
and its compare
method expects two Integer
objects. Integer::compareTo
can satisfy those expectations because it also expects two Integer
objects. The first is its receiver (the instance on which the method is to be called) and the second is the argument. With the new Java 8 syntax, the compiler translates one style to the other.
(compareTo
also returns an int
as required by Comparator#compare
.)
1
how doesInteger::compareTo
satisfy the signature ofComparator#compare
?compareTo(Integer)
vscompare(Integer, Integer)
?
– Andrew Tobilko
Apr 15 at 21:30
1
I disagree with "the compiler translates one style to the other". I think the compiler may validate a syntactic construct, but not translate it into another one. I am sure the compiler doesn't turn a method reference in a lambda. I didn't downvote your answer, btw.
– Andrew Tobilko
Apr 16 at 7:02
add a comment |
Integer
implements Comparable
by overriding compareTo
.
That overriden compareTo
, however, can be used in a way that satisfies and implements the Comparator
interface.
In its usage here
int value = intList.stream().max(Integer::compareTo).get();
it's translated to something like
int value = intList.stream().max(new Comparator<Integer>()
@Override
public int compare(Integer o1, Integer o2)
return o1.compareTo(o2);
).get();
A method reference (or lambda expression) must satisfy the signature of the corresponding functional interface's single abstract method and, in this case (Comparator
), compareTo
does.
The idea is that max
expects a Comparator
and its compare
method expects two Integer
objects. Integer::compareTo
can satisfy those expectations because it also expects two Integer
objects. The first is its receiver (the instance on which the method is to be called) and the second is the argument. With the new Java 8 syntax, the compiler translates one style to the other.
(compareTo
also returns an int
as required by Comparator#compare
.)
Integer
implements Comparable
by overriding compareTo
.
That overriden compareTo
, however, can be used in a way that satisfies and implements the Comparator
interface.
In its usage here
int value = intList.stream().max(Integer::compareTo).get();
it's translated to something like
int value = intList.stream().max(new Comparator<Integer>()
@Override
public int compare(Integer o1, Integer o2)
return o1.compareTo(o2);
).get();
A method reference (or lambda expression) must satisfy the signature of the corresponding functional interface's single abstract method and, in this case (Comparator
), compareTo
does.
The idea is that max
expects a Comparator
and its compare
method expects two Integer
objects. Integer::compareTo
can satisfy those expectations because it also expects two Integer
objects. The first is its receiver (the instance on which the method is to be called) and the second is the argument. With the new Java 8 syntax, the compiler translates one style to the other.
(compareTo
also returns an int
as required by Comparator#compare
.)
edited Apr 15 at 22:14
answered Apr 15 at 17:01
SaviorSavior
1,60921331
1,60921331
1
how doesInteger::compareTo
satisfy the signature ofComparator#compare
?compareTo(Integer)
vscompare(Integer, Integer)
?
– Andrew Tobilko
Apr 15 at 21:30
1
I disagree with "the compiler translates one style to the other". I think the compiler may validate a syntactic construct, but not translate it into another one. I am sure the compiler doesn't turn a method reference in a lambda. I didn't downvote your answer, btw.
– Andrew Tobilko
Apr 16 at 7:02
add a comment |
1
how doesInteger::compareTo
satisfy the signature ofComparator#compare
?compareTo(Integer)
vscompare(Integer, Integer)
?
– Andrew Tobilko
Apr 15 at 21:30
1
I disagree with "the compiler translates one style to the other". I think the compiler may validate a syntactic construct, but not translate it into another one. I am sure the compiler doesn't turn a method reference in a lambda. I didn't downvote your answer, btw.
– Andrew Tobilko
Apr 16 at 7:02
1
1
how does
Integer::compareTo
satisfy the signature of Comparator#compare
? compareTo(Integer)
vs compare(Integer, Integer)
?– Andrew Tobilko
Apr 15 at 21:30
how does
Integer::compareTo
satisfy the signature of Comparator#compare
? compareTo(Integer)
vs compare(Integer, Integer)
?– Andrew Tobilko
Apr 15 at 21:30
1
1
I disagree with "the compiler translates one style to the other". I think the compiler may validate a syntactic construct, but not translate it into another one. I am sure the compiler doesn't turn a method reference in a lambda. I didn't downvote your answer, btw.
– Andrew Tobilko
Apr 16 at 7:02
I disagree with "the compiler translates one style to the other". I think the compiler may validate a syntactic construct, but not translate it into another one. I am sure the compiler doesn't turn a method reference in a lambda. I didn't downvote your answer, btw.
– Andrew Tobilko
Apr 16 at 7:02
add a comment |
First trick: all instance methods actually take 1 additional implicit argument, the one you refer to as this
in method body. E.g.:
public final class Integer extends Number implements Comparable<Integer>
public int compareTo(/* Integer this, */ Integer anotherInteger)
return compare(this.value, anotherInteger.value);
Integer a = 10, b = 100;
int compareResult = a.compareTo(b);
// this actually 'compiles' to Integer#compareTo(this = a, anotherInteger = b)
Second trick: Java compiler can "transform" the signature of a method reference to some functional interface, if the number and types of arguments (including this
) satisfy:
interface MyInterface
int foo(Integer bar, Integer baz);
Integer a = 100, b = 1000;
int result1 = ((Comparator<Integer>) Integer::compareTo).compare(a, b);
int result2 = ((BiFunction<Integer, Integer, Integer>) Integer::compareTo).apply(a, b);
int result3 = ((MyInterface) Integer::compareTo).foo(a, b);
// result1 == result2 == result3
As you can see class Integer
implements none of Comparator
, BiFunction
or a random MyInterface
, but that doesn't stop you from casting the Integer::compareTo
method reference as those interfaces.
add a comment |
First trick: all instance methods actually take 1 additional implicit argument, the one you refer to as this
in method body. E.g.:
public final class Integer extends Number implements Comparable<Integer>
public int compareTo(/* Integer this, */ Integer anotherInteger)
return compare(this.value, anotherInteger.value);
Integer a = 10, b = 100;
int compareResult = a.compareTo(b);
// this actually 'compiles' to Integer#compareTo(this = a, anotherInteger = b)
Second trick: Java compiler can "transform" the signature of a method reference to some functional interface, if the number and types of arguments (including this
) satisfy:
interface MyInterface
int foo(Integer bar, Integer baz);
Integer a = 100, b = 1000;
int result1 = ((Comparator<Integer>) Integer::compareTo).compare(a, b);
int result2 = ((BiFunction<Integer, Integer, Integer>) Integer::compareTo).apply(a, b);
int result3 = ((MyInterface) Integer::compareTo).foo(a, b);
// result1 == result2 == result3
As you can see class Integer
implements none of Comparator
, BiFunction
or a random MyInterface
, but that doesn't stop you from casting the Integer::compareTo
method reference as those interfaces.
add a comment |
First trick: all instance methods actually take 1 additional implicit argument, the one you refer to as this
in method body. E.g.:
public final class Integer extends Number implements Comparable<Integer>
public int compareTo(/* Integer this, */ Integer anotherInteger)
return compare(this.value, anotherInteger.value);
Integer a = 10, b = 100;
int compareResult = a.compareTo(b);
// this actually 'compiles' to Integer#compareTo(this = a, anotherInteger = b)
Second trick: Java compiler can "transform" the signature of a method reference to some functional interface, if the number and types of arguments (including this
) satisfy:
interface MyInterface
int foo(Integer bar, Integer baz);
Integer a = 100, b = 1000;
int result1 = ((Comparator<Integer>) Integer::compareTo).compare(a, b);
int result2 = ((BiFunction<Integer, Integer, Integer>) Integer::compareTo).apply(a, b);
int result3 = ((MyInterface) Integer::compareTo).foo(a, b);
// result1 == result2 == result3
As you can see class Integer
implements none of Comparator
, BiFunction
or a random MyInterface
, but that doesn't stop you from casting the Integer::compareTo
method reference as those interfaces.
First trick: all instance methods actually take 1 additional implicit argument, the one you refer to as this
in method body. E.g.:
public final class Integer extends Number implements Comparable<Integer>
public int compareTo(/* Integer this, */ Integer anotherInteger)
return compare(this.value, anotherInteger.value);
Integer a = 10, b = 100;
int compareResult = a.compareTo(b);
// this actually 'compiles' to Integer#compareTo(this = a, anotherInteger = b)
Second trick: Java compiler can "transform" the signature of a method reference to some functional interface, if the number and types of arguments (including this
) satisfy:
interface MyInterface
int foo(Integer bar, Integer baz);
Integer a = 100, b = 1000;
int result1 = ((Comparator<Integer>) Integer::compareTo).compare(a, b);
int result2 = ((BiFunction<Integer, Integer, Integer>) Integer::compareTo).apply(a, b);
int result3 = ((MyInterface) Integer::compareTo).foo(a, b);
// result1 == result2 == result3
As you can see class Integer
implements none of Comparator
, BiFunction
or a random MyInterface
, but that doesn't stop you from casting the Integer::compareTo
method reference as those interfaces.
edited Apr 16 at 18:35
answered Apr 16 at 18:20
TairTair
3,1091329
3,1091329
add a comment |
add a comment |
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4
Integer::compareTo
does not return a Comparable - it is the short definition for: "Please compiler generate a matching implementation for the type that is needed (in this caseComparator
) and map the arguments to the specified function." In this case the function requires two "arguments" (this
and the one parameter ofcompareTo
) and the Comparator provides two arguments -> works.– Robert
Apr 15 at 18:37
6
""Please compiler, generate..." ... compilers always respond best to politeness and courtesy :-)
– scottb
Apr 15 at 18:45
1
Related: stackoverflow.com/questions/22561614/…
– Oleksandr
Apr 15 at 22:27